FDWK_3ed_Ch05_pp262-319

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Section 5.3 Definite Integrals and Antiderivatives 293 Extending the Ideas 52. Graphing Calculator Challenge If k 1, and if the average value of x k on 0, k is k, what is k Check your result with a CAS if you have one available. k 2.39838 53. Show that if Fx Gx on a, b, then Fb Fa Gb Ga. b F(x) dx a b G(x) dx→F(b) F(a) G(b) G(a) a Quick Quiz for AP* Preparation: Sections 5.1–5.3 You should solve the following problems without using a calculator. 1. Multiple Choice If b f x dx a 2b, then a f x 3 dx D b a (A) a 2b 3 (C) 4a – b (E) 5b – 3a 2. Multiple Choice The expression 1 1 2 0 20 (B) 3b – 3a (D) 5b – 2a 2 20 3 20 ... 2 0 20 is a Riemann sum approximation for 1 x (A) 0 20 dx 1 1 x (C) 2 0 0 20 dx (D) 1 20 (E) 2 1 0 20 0 x dx Answers to Section 5.3 Exercises 7. 0 1 0 sin (x 2 ) dx sin (1) 1 8. 1 8 dx 1 x 8 dx 1 9 dx 0 22 1 0 0 x 8 dx 3 9. 0 (b a) min f (x) b f (x) dx a 10. b f (x) dx (b – a) max f (x) 0 a 8 38. 1 0.5 1 7 0 1 x 4 dx 1 2 1 4 1 8 1 0.5 1 x 4 dx 1 7 4 9 1 1 68 0 1 x 4 dx 3 3 34 1 39. Yes; av( f ) b b f (x) dx, therefore b a a a f (x) dx av( f )(b a) b av( f ) dx. a 0 (B) B 1 x dx 0 0 1 x dx 3. Multiple Choice What are all values of k for which k 2 x2 dx 0 C (A) 2 (B) 0 (C) 2 (D) 2 and 2 (E) 2, 0, and 2 4. Free Response Let f be a function such that f(x) 6x 12. (a) Find f (x) if the graph of f is tangent to the line 4x y 5 at the point (0, 5). (b) Find the average value of f (x) on the closed interval [1, 1]. 4. (a) Since f (x) 6x 12, we have f (x) 3x 2 12x c for some constant c. The tangent line has slope 4 at (0, 5), so f (0) 4, from which we conclude that f (x) 3x 2 12x 4. This implies that f (x) x 3 6x 2 4x k for some constant k, and k must be 5 for the graph to pass through (0, 5). Thus f (x) x 3 6x 2 4x 5. (b) Average value 1 1 1 1 (1) (x 3 6x 2 4x 5)dx 1 2 x 4 2x 3 2x 2 5x 4 13. 1 40. (a) 300 miles (b) 8 hours (c) 37.5 mph total distance travelled (d) Average speed is for the whole trip. time d 1 d2 1 t t 2 2 d 1 d 2 t 1 t 2 1 41. Avg rate total amount released total time 2000 m 3 100 min 50 min 13 1 3 m3 /min 44. Let L(x) cx d. Then the average value of f on [a, b] is 1 av( f ) b b (cx d)dx a a 1 b a c b 2 db 2 c a 2 da 2 1 a 2 ) d(b a) b a c(b2 2 c(b a ) 2d 2 (ca d) (cb d ) 2 L(a) L(b) 2
294 Chapter 5 The Definite Integral 5.4 What you’ll learn about • Fundamental Theorem, Part 1 • Graphing the Function x a ft dt • Fundamental Theorem, Part 2 • Area Connection • Analyzing Antiderivatives Graphically . . . and why The Fundamental Theorem of Calculus is a triumph of mathematical discovery and the key to solving many problems. Sir Isaac Newton (1642–1727) Sir Isaac Newton is considered to be one of the most influential mathematicians of all time. Moreover, by the age of 25, he had also made revolutionary advances in optics, physics, and astronomy. Fundamental Theorem of Calculus Fundamental Theorem, Part 1 This section presents the discovery by Newton and Leibniz of the astonishing connection between integration and differentiation. This connection started the mathematical development that fueled the scientific revolution for the next 200 years, and is still regarded as the most important computational discovery in the history of mathematics: The Fundamental Theorem of Calculus. The Fundamental Theorem comes in two parts, both of which were previewed in Exploration 2 of the previous section. The first part says that the definite integral of a continuous function is a differentiable function of its upper limit of integration. Moreover, it tells us what that derivative is. The second part says that the definite integral of a continuous function from a to b can be found from any one of the function’s antiderivatives F as the number Fb Fa. THEOREM 4 The Fundamental Theorem of Calculus, Part 1 If f is continuous on a, b, then the function Fx x f t dt has a derivative at every point x in a, b, and d F d dx d x x a a f t dt f x. Proof The geometric exploration at the end of the previous section contained the idea of the proof, but it glossed over the necessary limit arguments. Here we will be more precise. Apply the definition of the derivative directly to the function F. That is, d F lim Fx h Fx dx h→0 h lim h→0 xh a f t dt x f t dt a h lim h→0 lim h→0 [ h xh 1 h f t dt] . x The expression in brackets in the last line is the average value of f from x to x h. We know from the Mean Value Theorem for Definite Integrals (Theorem 3, Section 5.3) that f, being continuous, takes on its average value at least once in the interval; that is, 1 xh h f t dt f c for some c between x and x h. x xh x f t dt Rules for integrals, Section 5.3
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FDWK_3ed_Ch05_pp262-319

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